Simplify and expand the following expression: $ \dfrac{3n}{n - 3}+\dfrac{2n}{3n - 3} $
Answer: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(n - 3)(3n - 3)$ Multiply the first term by $\dfrac{3n - 3}{3n - 3}$ $ \begin{align*} \dfrac{3n}{n - 3} \times \dfrac{3n - 3}{3n - 3} & = \dfrac{(3n)(3n - 3)}{(n - 3)(3n - 3)} \\ & = \dfrac{9n^2 - 9n}{(n - 3)(3n - 3)}\end{align*} $ Multiply the second term by $\dfrac{n - 3}{n - 3}$ $ \begin{align*} \dfrac{2n}{3n - 3} \times \dfrac{n - 3}{n - 3} & = \dfrac{(2n)(n - 3)}{(3n - 3)(n - 3)} \\ & = \dfrac{2n^2 - 6n}{(3n - 3)(n - 3)}\end{align*} $ Now we have: $ = \dfrac{9n^2 - 9n}{(n - 3)(3n - 3)} + \dfrac{2n^2 - 6n}{(3n - 3)(n - 3)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{9n^2 - 9n + 2n^2 - 6n}{(n - 3)(3n - 3)} $ $ = \dfrac{11n^2 - 15n}{(n - 3)(3n - 3)}$ Expand the denominator: $ = \dfrac{11n^2 - 15n}{3n^2 - 12n + 9}$